0

I have a notation question: Which partial derivative notation for $\partial_j f_{,i}$ is correct: $f_{,ji}$ or $f_{,ij}$?

I think $f_{,ji}$ is correct.

Best regards

Bruno Lobo
  • 496
  • Why would you mix notations like that in the first place? It's a matter of if you read it as $(\partial_j f){,i}$, or $\partial_j(f{,i})$. – Mr. Brown Jan 10 '23 at 16:17
  • It is an equivalence relation on notation space. Does it make sense? It may be useful on Einstein's notation usage – Bruno Lobo Jan 10 '23 at 16:18
  • Like I said, you're just inviting trouble for yourself. This notation is bad precisely because it can be interpreted two ways if you do not put in parentheses. – Mr. Brown Jan 10 '23 at 16:20
  • It is not an invitation if my surname is "trouble". – Bruno Lobo Jan 10 '23 at 16:20
  • In my opinion, these you propose are different situation, which have different notations: $(f_{j})i$ is not in Einstein's notation, the correct form is $f{,j}^i$. – Bruno Lobo Jan 10 '23 at 16:23
  • Even for Einstein notation, it's better to stick to one. Personally, I find commas clearer because of products. By this I mean $\partial_i fg$ vs $f_i g$. Is there a specific instance where you have a strong case for the mixed notation? – Mr. Brown Jan 10 '23 at 16:24
  • Stack-bot asked for us to take a room for discussion. Will you join? – Bruno Lobo Jan 10 '23 at 16:25
  • No, I'm busy right now. I'll end with this: $f^i_,j$ is not the same as $f_{,ij}$! The first denotes taking the $j$th derivative of the function $f$ after projected onto the $i$th component. The second denotes taking the $i$ then $j$ partial derivative of the function $f$. – Mr. Brown Jan 10 '23 at 16:27
  • Ok. Your latter explanation clarifies the question as $\partial_j f_{,i} = \partial_j f_{,ij}$. I am inclined to believe this is not correct because it must be consistent to left-right order on operator $\frac{\partial}{\partial x^j , \partial x^i}$-application – Bruno Lobo Jan 10 '23 at 16:31

1 Answers1

2

$\partial_j f_{,i}=\frac{\partial}{\partial x_j}\left(\frac{\partial f}{\partial x_i}\right)$

Therefore $f_{,ij}$

ryaron
  • 1,091